Open Ball in Euclidean Plane is Interior of Circle

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Theorem

Let $\R^2$ be the real number plane with the usual (Euclidean) metric.

Let $x = \tuple {x_1, x_2} \in \R^2$ be a point in $\R^2$.

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.


Then $\map {B_\epsilon} x$ is the interior of the circle whose center is $x$ and whose radius is $\epsilon$.


Proof

Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.

Let $y = \tuple {y_1, y_2} \in \map {B_\epsilon} x$.

Then:

\(\ds y\) \(\in\) \(\, \ds \map {B_\epsilon} x \, \) \(\ds \)
\(\ds \leadstoandfrom \ \ \) \(\ds \map d {y, x}\) \(<\) \(\, \ds \epsilon \, \) \(\ds \) Definition of Open $\epsilon$-Ball
\(\ds \leadstoandfrom \ \ \) \(\ds \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2}\) \(<\) \(\, \ds \epsilon \, \) \(\ds \) Definition of Real Number Plane with Euclidean Metric
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2\) \(<\) \(\, \ds \epsilon^2 \, \) \(\ds \)

But from Equation of Circle:

$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 = \epsilon^2$

is the equation of a circle whose center is $\tuple {x_1, x_2}$ and whose radius is $\epsilon$.


The result follows by definition of interior and Open Ball of Point Inside Open Ball.

$\blacksquare$


Sources

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